November 2019 Paper 1 Higher

$108 = 2^2 \times 3^3$

$120 = 2^3 \times 3 \times 5$

Take the highest power of each prime:

LCM $= 2^3 \times 3^3 \times 5 = 8 \times 27 \times 5 = 1080$

Women: $60 \div 2 = 30$

Men: $30 \div 3 = 10$

Children: $60 - 30 - 10 = 20$

Children : men $= 20 : 10 = 2 : 1$, so $n = 2$

$1\frac{3}{4} = \frac{7}{4}$ and $1\frac{1}{3} = \frac{4}{3}$

$\frac{7}{4} \times \frac{4}{3} = \frac{28}{12} = \frac{7}{3} = 2\frac{1}{3}$

1. With centre $P$, draw an arc cutting $CD$ at two points.
2. From each of these points, draw arcs of equal radius that cross on the other side of $CD$.
3. Join $P$ to where the arcs cross. This line meets $CD$ at right angles.

Leave all arcs showing.

Angle $ACB = 180 - 75 - 51 = 54°$

Split 54 in the ratio 2 : 1: one part $= 54 \div 3 = 18°$

So $ACD = 18°$ and $DCB = 36°$

In triangle $BDC$: angle $BDC = 180 - 51 - 36 = 93°$

Red: $4 \times 5 = 20$ kg

Blue: $5 \times 9 = 45$ kg

Total $= 20 + 45 + 6 = 71$ kg

Mean $= 71 \div 10 = 7.1$ kg

$7.1 > 7$, so Donna is not correct.

(a) Multiply the powers: $(p^2)^5 = p^{2 \times 5} = p^{10}$

(b) Divide the numbers and subtract the powers:

$12 \div 6 = 2$, $x^{7-3} = x^4$, $y^{3-1} = y^2$

Answer: $2x^4y^2$

Distance $PQ = 12 \times 1.5 = 18$ km

On the drawing: $18 \div 4 = 4.5$ cm

From $P$, measure $70°$ clockwise from north and mark $Q$ 4.5 cm along that line.

(i) Measure $QL$ in cm and multiply by 4 (about 20 to 23 km).

(ii) At $L$, measure clockwise from north round to $Q$: about $317°$ to $330°$.

(a) 18 minutes $= \frac{18}{60} = 0.3$ hours

Distance $= 72 \times 0.3 = 21.6$ km

(b) $72$ km/h $= 72\,000$ m in $3600$ s

$72\,000 \div 3600 = 20$ m/s. The speeds are the same, so David is not correct.

(a) Plot each cumulative frequency at the end of its interval and join with a smooth curve or straight lines, starting at (20, 0).

(b) LQ at $cf = 10$: about 45; UQ at $cf = 30$: about 70

IQR $\approx 70 - 45 = 25$ minutes

(c) At 90: $cf \approx 36.5$; at 50: $cf \approx 15$

$36.5 - 15 = 21.5$, so probability $\approx \frac{21.5}{40} \approx 0.54$

In the sample, $\frac{5}{30} = \frac{1}{6}$ of the counters are yellow.

So 12 yellow counters is about $\frac{1}{6}$ of the bag.

$p \approx 12 \times 6 = 72$

Multiplying both sides by 2 means every term is doubled:

$2T = q + 10$

Spencer left the 5 as 5. The correct result is $q = 2T - 10$.

(a) Common denominator $3x(x+1)$:

$\frac{5 \times 3x}{3x(x+1)} + \frac{2(x+1)}{3x(x+1)} = \frac{15x + 2x + 2}{3x(x+1)} = \frac{17x+2}{3x(x+1)}$

(b) $(x+y)$ is a common factor:

$(x+y)^2 + 3(x+y) = (x+y)(x+y+3)$

$\frac{1}{2}(x+4)(x-2) = 27.5$

$(x+4)(x-2) = 55 \Rightarrow x^2 + 2x - 8 = 55$

$x^2 + 2x - 63 = 0 \Rightarrow (x+9)(x-7) = 0$

Lengths are positive, so $x = 7$

Shortest side $= x - 2 = 5$ cm

Let $x = 0.4181818\ldots$

$1000x = 418.181818\ldots$ and $10x = 4.181818\ldots$

$990x = 414$

$x = \frac{414}{990} = \frac{23}{55}$

(a) $\frac{22}{\sqrt{11}} \times \frac{\sqrt{11}}{\sqrt{11}} = \frac{22\sqrt{11}}{11} = 2\sqrt{11}$

(b) Multiply by $\frac{2\sqrt{3}+1}{2\sqrt{3}+1}$:

Top: $\sqrt{3}(2\sqrt{3}+1) = 6 + \sqrt{3}$

Bottom: $(2\sqrt{3})^2 - 1^2 = 12 - 1 = 11$

So it equals $\frac{6+\sqrt{3}}{11}$, with $a = 6$ and $b = 11$.

Area ratio $4 : 9$, so length ratio $\sqrt{4} : \sqrt{9} = 2 : 3$

Volume ratio $2^3 : 3^3 = 8 : 27$

$27 \div 8 = 3.375$

Three fills are not enough, so Tyler fills A 4 times.

(a) $f(3) = 2 \times 3^3 - 4 = 54 - 4 = 50$, so $f^{-1}(50) = 3$

(b) $hg(x) = (x+2)^2 = x^2 + 4x + 4$

$x^2 + 4x + 4 = 3x^2 + x - 1$

$0 = 2x^2 - 3x - 5 = (2x - 5)(x + 1)$

$x = 2.5$ or $x = -1$

$9^{-\frac{1}{2}} = 3^{-1}$ and $27^{\frac{1}{4}} = 3^{\frac{3}{4}}$

$3^{-1} = 3^{\frac{3}{4} - (x+1)}$

$-1 = \frac{3}{4} - x - 1$

$x = \frac{3}{4}$

(a) $f(x+1)$ moves the graph 1 unit left; $-3$ moves it 3 units down.

So each corner moves by $\begin{pmatrix}-1\\-3\end{pmatrix}$, for example $(-2, 1) \to (-3, -2)$ and $(1, 4) \to (0, 1)$.

(b) $f(-x)$ reflects in the $y$-axis, so $x$ changes sign: $B = (2, 1)$

$y$-intercept: $x = 0$ gives $y = -5$

Complete the square: $2x^2 - 8x - 5 = 2(x - 2)^2 - 8 - 5 = 2(x-2)^2 - 13$

Turning point: $(2, -13)$

$x$-intercepts: $2(x-2)^2 = 13 \Rightarrow x = 2 \pm \sqrt{\frac{13}{2}}$

Sketch a U shape through these points.

Triangle $AED$ is equilateral, so each angle is $60°$.

$\angle ACB = \angle ADB = 60°$ and $\angle DBC = \angle DAC = 60°$ (angles in the same segment).

So triangle $BEC$ is equilateral and $\angle ACB = \angle DBC$.

$AC = AE + EC = DE + EB = DB$ (sides of equilateral triangles).

$BC$ is common to both triangles.

So triangles $ABC$ and $DCB$ are congruent (SAS).